Asked by Kaelyn
Solve the following equations
7f+f^2=60
6x^2-x-15=0
6(y-3)^2-(y-3)-15=0
7f+f^2=60
6x^2-x-15=0
6(y-3)^2-(y-3)-15=0
Answers
Answered by
Steve
always rearrange things into descending powers. Set equal to zero because after you factor it, you know that the product can be zero only if one of the factors is zero.
f^2+7f-60 = 0
(f+12)(f-5) = 0
f = -12 or 5
6x^2-x-15 = 0
(2x+3)(3x-5) = 0
x = -3/2 or 5/3
6(y-3)^2 - (y-3) - 15 = 0
(2(y-3)+3)(3(y-3)-5) = 0
one way: y-3 = -3/2 or 5/3
That means y = 3/2 or 14/3
other way:
(2y-6+3)(3y-9-14) = 0
(2y-3)(3y-14) = 0
y = 3/2 or 14/3
f^2+7f-60 = 0
(f+12)(f-5) = 0
f = -12 or 5
6x^2-x-15 = 0
(2x+3)(3x-5) = 0
x = -3/2 or 5/3
6(y-3)^2 - (y-3) - 15 = 0
(2(y-3)+3)(3(y-3)-5) = 0
one way: y-3 = -3/2 or 5/3
That means y = 3/2 or 14/3
other way:
(2y-6+3)(3y-9-14) = 0
(2y-3)(3y-14) = 0
y = 3/2 or 14/3
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.